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Math Help - Two Continuity problems

  1. #16
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    Re: Two Continuity problems

    Plato asked if you could prove f(0)= 0, then gave a proof himself. You responded to that with
    i understand , u mean if F(x) = x-2 then f(0) <0 , and if F(x)=x+2 then f(0) > 0 , but i cant get how to use this to solve my question !!
    No, Plato meant nothing of the sort- he certainly never said anything about the function "F(x)= x- 2" because that is NOT an odd function! Further he was not talking about two different functions which is what you are doing.

    A function is odd if, for all x, f(-x)= -f(x). That is not true for F(x)= x- 2: Taking x= 3, F(3)= 3- 2= 1. Taking x= -3, F(-3)= -3- 2= -5. F(-x)\ne-F(x).

    What Plato was saying was- start from f(-x)= -f(x) and let x= 0.

    To show that a function, f, is continuous at x= a, we would have to prove that \lim_{x\to a} f(x)= f(a). Here, we have a= 0 and we know that f(0)= 0 so that reduces to \lim_{x\to 0} f(x)= 0. We are also told that f is "right continuous at x= 0" which means \lim_{x\to 0^+} f(x)= 0 (that is the limit as we approach 0 only from the positive numbers). Now you know that f(-x)= -f(x).
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  2. #17
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    Re: Two Continuity problems

    Quote Originally Posted by HallsofIvy View Post
    Plato asked if you could prove f(0)= 0, then gave a proof himself. You responded to that with
    No, Plato meant nothing of the sort- he certainly never said anything about the function "F(x)= x- 2" because that is NOT an odd function! Further he was not talking about two different functions which is what you are doing.

    A function is odd if, for all x, f(-x)= -f(x). That is not true for F(x)= x- 2: Taking x= 3, F(3)= 3- 2= 1. Taking x= -3, F(-3)= -3- 2= -5. F(-x)\ne-F(x).

    What Plato was saying was- start from f(-x)= -f(x) and let x= 0.

    To show that a function, f, is continuous at x= a, we would have to prove that \lim_{x\to a} f(x)= f(a). Here, we have a= 0 and we know that f(0)= 0 so that reduces to \lim_{x\to 0} f(x)= 0. We are also told that f is "right continuous at x= 0" which means \lim_{x\to 0^+} f(x)= 0 (that is the limit as we approach 0 only from the positive numbers). Now you know that f(-x)= -f(x).

    Thx for ur help , this was very useful .. and thxx to plato for trying to help first
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